一个长度为10000的字符串，写一个算法，找出最长的重复子串，如abczzacbca,结果是bc

[gb2312iso]

一个长度为10000的字符串，写一个算法，找出最长的重复子串，如abczzacbca,结果是bc。求解答。。。

[不关橙猫猫事的哦]

abczzacbca,为什么zz不算啊？zz也是长2啊。。。应该是重复最多的吧。。

[ousyuryu]

10000个字符串搞组合？怎么想出来的啊

[不关橙猫猫事的哦]

[Quote=引用 2 楼 ousyuryu 的回复:]

10000个字符串搞组合？怎么想出来的啊
[/Quote]
貌似是面试题。。考hashmap的。。

[lhjtianji]

最长的当然是bc了，hashmap已经忘得差不多了，观望

[剑神一笑]

稍微优化了下:

public static void main(String[] args) {
		long startTime = System.currentTimeMillis();
		String s = "abczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbca"
				  +"abczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbcaabczzacbca"
				  +"";
		test1(s);
		long endTime = System.currentTimeMillis();
		System.out.println("花费时间为："+((double)(endTime-startTime))/1000+"秒");
	}
	
	public static void test1(String s){
		String temp = null;
		int tempLen = 0;
		char[] c = s.toCharArray();
		Map<String, Integer> m = new HashMap<String, Integer>();
		for (int i = 0; i < c.length - 1; i++) {
			temp = c[i] + "" + c[i + 1];
			tempLen = i + 1;
			while (s.indexOf(temp)!=s.lastIndexOf(temp)) {
				if (m.containsKey(temp.toString())) {
					m.put(temp, m.get(temp) + 1);
				} else {
					m.put(temp, 1);
				}
				tempLen++;
				if (tempLen >= c.length) {
					break;
				} else {
					temp += c[tempLen];
				}
			}
		}
		System.out.println("s的长度为："+s.length());
		Map<String, Integer> m1 = new HashMap<String, Integer>();
		// 去掉值为1的子串
		for (Entry<String, Integer> e : m.entrySet()) {
			if (e.getValue() != 1) {
				m1.put(e.getKey(), e.getValue());
			}
		}
		if(m1.size()==0){
			System.out.println("没有最长子串！");
		}else{
			// 取出最大的子串
			int max = 0;
			for (Entry<String, Integer> e : m1.entrySet()) {
				if(e.getValue()>max){
					max = e.getValue();
				}
			}
			StringBuffer sb = new StringBuffer();
			for (Entry<String, Integer> e : m1.entrySet()) {
				if(e.getValue()==max){
					sb.append(e.getKey());
					sb.append(",");
				}
			}
			System.out.println("最长子串为："+sb.substring(0, sb.length()-1)+"  长度为："+max);
		}
	}

s的长度为：2000
最长子串为：bc  长度为：400
花费时间为：19.766秒

s的字符串为2000需要20秒 3000就会内存溢出 需要加大JVM内存

[gb2312iso]

[Quote=引用 1 楼 bill0605030109 的回复:]

abczzacbca,为什么zz不算啊？zz也是长2啊。。。应该是重复最多的吧。。
[/Quote]

找出最长的重复子串 zz 只出现过一次 而bc出现过两次

[qq497525725]

这个 真不会。

[qq497525725]

不会 围观。

[剑神一笑]

public static void main(String[] args) {
		String s = "abczzacbca";
		String temp = null;
		int tempLen = 0;
		char[] c = s.toCharArray();
		Map<String, Integer> m = new HashMap<String, Integer>();
		for (int i = 0; i < c.length - 1; i++) {
			temp = c[i] + "" + c[i + 1];
			tempLen = i + 1;
			while (s.indexOf(temp.toString()) != tempLen) {
				if (m.containsKey(temp.toString())) {
					m.put(temp, m.get(temp) + 1);
				} else {
					m.put(temp, 1);
				}
				tempLen++;
				if (tempLen >= c.length) {
					break;
				} else {
					temp += c[tempLen];
				}
			}
		}
		Map<String, Integer> m1 = new HashMap<String, Integer>();
		// 去掉值为1的子串
		for (Entry<String, Integer> e : m.entrySet()) {
			if (e.getValue() != 1) {
				m1.put(e.getKey(), e.getValue());
			}
		}
		if(m1.size()==0){
			System.out.println("没有最长子串！");
		}else{
			// 取出最大的子串
			int max = 0;
			for (Entry<String, Integer> e : m1.entrySet()) {
				if(e.getValue()>max){
					max = e.getValue();
				}
			}
			StringBuffer sb = new StringBuffer();
			for (Entry<String, Integer> e : m1.entrySet()) {
				if(e.getValue()==max){
					sb.append(e.getKey());
					sb.append(",");
				}
			}
			System.out.println("最长子串为："+sb.substring(0, sb.length()-1)+"  长度为："+max);
		}
	
	}

没优化的
不知道对10000的字符串有没有效果
1000一下的应该是没问题的

[getserved]

public void test3(){
		String str = "abczzacbca";
		int strlen = str.length();
		int findlen = 2;
		int pointer = 0;
		for(int i = 0; i < strlen - 2; i++){
			String toFind = str.substring(pointer, pointer+findlen);
			for(int j = pointer+2; j < strlen-2;j++){
				String comp = str.substring(j,j+findlen);
				System.out.println(find(str, toFind, comp, pointer, j,findlen));
			}
			pointer++;
		}
	}
	
	public String find(String str, String toFind, String comp, int pointer, int compPointer, int findlen){
		String found = "";
		String lastFound = toFind;
		if(toFind.equals(comp)){
			findlen++;
			toFind = str.substring(pointer,pointer+findlen);
			comp = str.substring(compPointer+findlen);
			if(pointer < compPointer){
				found = find(str, toFind, comp, pointer, compPointer,findlen);
				if(found == ""){
					found = lastFound;
				}
			}
		}
		return found;
	}